# GR matter-free equations and Schwarzschild geometry

I am reading some lecture notes on General relativity (undergraduate level) and I do not understand a sequence of statements about the topics in the title.

After stating that the for matter-free space the components of the Ricci tensor vanish, they go on to say that non-trivial solutions to these equations would represent the propagation of gravitational waves through otherwise empty space. Then, they proceed to using that equation to derive the Schwarzschild metric, assuming a stationary and radially symmetric situation. After getting the expression for the line element they say that even though the Ricci tensor is zero for $r>0$ this does not mean that it is zero identically and that the Schwarzschild metric is the metric of a point mass at the origin of spacial coordinate.

I don't understand why this is the case and I thought that we were under the assumption that we were in empty space?

Furthermore, they then state the Birkhoff's theorem and say that it means that a radially pulsating star cannot emit gravitational waves.

Then when investigating black hole geometry they assume that the source of the Schwarzschild metric is some massive object within its Schwarzschild radius.

This raises several questions for me. Firstly, I thought the Schwarzschild metric was for empty space so I don't understand how we can be talking about its source being a massive object. Secondly, I thought that the Schwarzschild metric was a non-trivial solution to the matter free equations and according to their initial statements this would represent gravitational waves. But this conflicts with Birkhoff's theorem. Could you help me identify what concept(s) I am misunderstanding?

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There are essentially two sources of problem from what I see: first concerning solutions of Einstein Equations in vacuum, and secondly concerning Birkhoff's theorem in the case of spherically symmetrical solutions with matter. Let's tackle them one at a time.

1)The curvature of a manifold is described by the Riemann tensor $R_{\mu\nu\lambda\rho}$, that can be canonically separated in a trace part, the Ricci tensor $R_{\mu\nu}$, and a trace-free part, the Weyl tensor $C_{\mu\nu\lambda\rho}$. The Einstein Equations relate the Ricci tensor with the matter distribution by the energy-momentum tensor $T_{\mu\nu}$ in the form

$R_{\mu\nu}=8\pi G(T_{\mu\nu}-1/2g_{\mu\nu}T)$.

You're interested in vacuum solutions, i.e. $T_{\mu\nu}=0$, which imply Ricci-flatness $R_{\mu\nu}=0$. Now, what about the Weyl tensor? It is determined by the second Bianchi identity $R_{\mu\nu [\lambda\rho;\sigma]}=0$, or in other words it satisfies a non-linear differential equation. Now the lectures you mention state that "non-trivial solutions to these equations would represent the propagation of gravitational waves through otherwise empty space". That is incorrect, and the Schwarzschild metric is a clear example of Ricci flat spacetime without waves. In general the Petrov Classification of the Weyl tensor is a way to determine if a solution possess or not gravitational waves. In particular all solutions of Petrov type D are everywhere vacuum metrics without gravitational waves, of which Schwarzschild is an example, though not the only one. The Petrov classification is a bit high level for an introduction to general relativity, but maybe the link at wikipedia can give you the flavor of the idea.

Regarding the assertion "even though the Ricci tensor is zero for $r>0$ this does not mean that it is zero identically and that the Schwarzschild metric is the metric of a point mass at the origin of spacial coordinate" this is plainly wrong. The only sense I can make of this phrase is what is contained in Jerry Schirmer's answer, where he very properly asserts that is strictly heuristical and should not be taken seriously. The reason is that unlike electromagnetism, general relativity involves non-linear equations for the metric, and that does not accommodates easily distributional sources, as Dirac delta one, otherwise the metric itself possibly won't have all the usual derivatives, which could lead to violating the Bianchi identities and consequently energy-momentum conservation.

So the lectures are wrong and you are right, you're working in empty space. Furthermore Schwarzschild is a non-trivial solution in vacuum without gravitational waves.

2)In second case the lectures look at the problem of a spherically symmetrical distribution of matter contained in given finite radius, let's call it $R_s$. Now for $r<R_s$ we have non-zero energy-momentum tensor and Einstein Equation must be solved accordingly. For instance, in the simple case of isotropic matter you have TOV equation. For $r>R_s$ you are in empty space, and therefore holds Birkhoff's theorem, which is a local assertion that the metric of static and spherically symmetric solutions must be Schwarzschild. The point you're missing is that the result is local and therefore ignores what's happening inside.

So you know that for radius below $R_s$ you must solve for whatever is the matter you have, outside is just Schwarzschild and at $r=R_s$ you must impose continuity of the metric, giving you the boundary condition to solve the diff. equation inside the star. Reiterating what I've previously wrote, since Birkhoff's theorem is local it does not care what is the solution inside the star. In particular it is irrelevant if the solution inside is time dependent (like an oscillation), as long as it remains spherically symmetric. So no conflict with the theorem.

The same holds for electromagnetism, as in a spherical distribution of charge which oscillates radially cannot emit radiation, a consequence of Gauss' law discussed in the majority of electromagnetism textbooks.

ADDENDUM: If you're having trouble with the lectures notes you're reading, and from what you say it is not your fault if it states slightly incorrect things, I suggest you try to put your hands on a copy of Hartle's book, it is the best is my opinion at undergrad level. If you don't have access to it then there's the freely available notes from Chrúsciel, which are great, albeit somewhat more high-level mathematically speaking.

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Thank you very much for your comprehensive answer! I will check out the notes you referenced. I am supporting the lecture notes with other resources but the approaches in textbooks are often so different in level or technique that it is tricky. –  Student May 17 at 21:58
In fact it is common knowledge that different GR textbooks diverge widely in approach. Robert Wald has written an article regarding the teaching of GR with a list of textbooks with comments on their approaches arxiv.org/abs/gr-qc/0511073. Maybe it will help you decide for a book, and allow you to compare different ones that have similar styles so you can support one with the other. –  cesaruliana May 18 at 19:58

NOTE: this is merely a heuristic. A rigorous proof of the mass in the schwarzschild spacetime involves taking the ADM or Bondi mass, or at least using covariant integrals. This is, IMO, slightly beyond the scope of undergraduate relativity.

The easiest way to see this is to note that (I'm going to abandon general tensor notation, because it confuses the issue in this case):

$$G_{tt} = 8\pi T_{tt} = 8\pi \rho$$

If you use the standard expression for the Schwarzschild metric: $g_{ab} = {\rm diag}(-(1-\frac{2M}{r}), \frac{1}{1-\frac{2M}{r}},r^{2},r^{2}\sin^{2}\theta)$, you can show that $G_{tt} = \nabla^{2}\frac{M}{r}$. It turns out that this expression is equal to $8\pi M\delta^{3}(r)$, by the properties of the delta function, we therefore have $\rho = 0$ for all $r\neq 0$, but $\int d^{3}x\, \rho = M$

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I don't understand why this is the case and I thought that we were under the assumption that we were in empty space?

It's a bit misleading. For the same reason we choose to solve Laplace's equation in spherical coordinates for the electric potential when there is charge only at the spatial origin, we choose to solve the vacuum equations for a spherically symmetric static spacetime in the case that there is mass-energy only at the spatial origin.

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The Schwarzschild metric applies in the region outside any spherically symmetric, massive object, to no closer than the Schwarzschild radius. You might recall that, in electromagnetism, a spherically symmetric charge distribution will look like a point charge in the exterior. The same is true in general relativity for spherically symmetric mass distributions.

The tendency to refer to the stress-energy tensor of a black hole as being "vacuum" is, in my opinion, unfortunate. In electromagnetism, we have no problem thinking of point charges as delta functions. The source term for a black hole could be thought of the same way. I believe this isn't done because differential geometry attacks the problem by excluding a point (a "punctured" domain) instead, so in that perspective, the stress-energy tensor is zero everywhere on the domain--it's just that the domain doesn't include the origin.

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Actually, in the extended spacetime, the singularity of Schwarzschild is a spacelike line, and the Kerr/Nordstrom metric is a timelike line. This stuff turns out to be more subtle than the case in electromagnetism, so, despite what I said above about delta functions, it's not 100% correct to think of the schwarzschild matter distribution as a delta function. –  Jerry Schirmer May 17 at 19:01